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CBSE Class 10 Answered

The sum of the first seven terms of an AP is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the AP.
Asked by Devika Mohan | 12 Mar, 2015, 11:08: PM
answered-by-expert Expert Answer

Let space straight a space be space the space first space term space and space straight d space be space the space comommon space difference space of space the space AP. The space nth space term space of space AP space is space given space by straight a subscript straight n equals straight a plus open parentheses straight n minus 1 close parentheses straight d The space sum space of space first space straight n space terms space of space AP space is space given space by straight S subscript straight n equals straight n over 2 open square brackets 2 straight a plus open parentheses straight n minus 1 close parentheses straight d close square brackets Given space that space the space sum space of space the space first space seven space terms space of space the space AP space is space 182. straight S subscript 7 equals 7 over 2 open square brackets 2 straight a plus open parentheses 7 minus 1 close parentheses straight d close square brackets 182 equals 7 over 2 open square brackets 2 straight a plus 6 straight d close square brackets 26 equals straight a plus 3 straight d......... left parenthesis 1 right parenthesis  Fourth space term space of space AP space is space given space by space straight a subscript 4 equals straight a plus open parentheses 4 minus 1 close parentheses straight d straight a subscript 4 equals straight a plus 3 straight d straight a subscript 4 equals 26 space space space space left parenthesis from space space space left parenthesis 1 right parenthesis right parenthesis  Sventeenth space term space of space AP space is space given space by straight a subscript 17 equals straight a plus 16 straight d  4 th space and space the space 17 th space space terms space are space in space the space ratio space 1 space colon space 5  straight a subscript 4 over straight a subscript 17 equals 1 fifth 26 cross times 5 equals straight a subscript 17 straight a subscript 17 equals 130  Consider comma straight a subscript 17 minus straight a subscript 4 equals straight a plus 16 straight d minus open parentheses straight a plus 3 straight d close parentheses 130 minus 26 equals 13 straight d 104 equals 13 straight d straight d equals 8  Put space straight d equals 8 space in space equation space left parenthesis 1 right parenthesis space we space get 26 equals straight a plus 24 straight a equals 2  so space the space AP space is space 2 comma space 10 comma space 18 comma space 26 comma 34 comma space...........

Answered by | 13 Mar, 2015, 12:22: PM
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